Is there an elegant way to cycle through a list N times via iteration (like itertools.cycle but limit the cycles)?

Question

I\'d like to cycle through a list repeatedly (N times) via an iterator, so as not to actually store N copies of the list in memory. Is there a built-in or elegant way to do

Answer 1

itertools.chain.from_iterable(iter(L) for x in range(N))

Answer 2

For the special case where you need to iterate over a list several times, this is not too bad.

It does create a list of n references to my_list, so if n is very large it is better to use Darthfelt's answer

>>> import itertools as it
>>> it.chain(*[my_list]*n)

Answer 3

All the other answers are excellent. Another solution would be to use islice. This allows you to interrupt the cycle at any point:

>>> from itertools import islice, cycle
>>> l = [1, 2, 3]
>>> list(islice(cycle(l), len(l) * 3))
[1, 2, 3, 1, 2, 3, 1, 2, 3]
>>> list(islice(cycle(l), 7))
[1, 2, 3, 1, 2, 3, 1]

Answer 4

import itertools
itertools.chain.from_iterable(itertools.repeat([1, 2, 3], 5))

Itertools is a wonderful library. :)

Answer 5

You said that you don't want to write your own generator, but a generator expression would probably be the easiest and most efficient way to accomplish what you're after. It doesn't require any function calls or importing of any modules. itertools is a great module, but maybe not necessary in this case?

some_list = [1, 2, 3]
cycles = 3
gen_expr = (elem for _ in xrange(cycles) for elem in some_list)

or just

(elem for _ in xrange(3) for elem in [1, 2, 3])

or

for elem in (e for _ in xrange(3) for e in [1, 2, 3]):
    print "hoo-ray, {}!".format(elem)

Answer 6

@Darthfett's answer is documented as an itertools recipes:

from itertools import chain, repeat

def ncycles(iterable, n):
    "Returns the sequence elements n times"
    return chain.from_iterable(repeat(tuple(iterable), n))


list(ncycles(["a", "b"], 3))
# ['a', 'b', 'a', 'b', 'a', 'b']

For convenience, I add that the more_itertools library implements this recipe (and many others) for you:

import more_itertools as mit

list(mit.ncycles(["a", "b"], 3))
# ['a', 'b', 'a', 'b', 'a', 'b']